Define sequences $$a_n = \sum_{k \in \mathbb{Z}}\left(\dfrac{\sin(k)}{k}\right)^{n}, b_n = \int_{0}^{\infty}\left(\dfrac{\sin(x)}{x}\right)^{n}dx.$$I am trying to see if there is a relation between these two sequences by studying sequence $c_n = a_n-2b_n$. From [1,2,3,4,5], sequence $b_n$ is completely understood, as $$ b_n=\frac{\pi}{2^n (n-1)!} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}$$and the first few values are $$ b_1 = b_2 = \frac{\pi}{2}, b_3=\frac{3\pi}{8}, b_4 = \frac{\pi}{3}, b_5 = \frac{115\pi}{384}, b_6 = \frac{11\pi}{40}.$$Now coming to the other sequence, [6,7], $a_n$ is also completely known,$$ a_n= \frac{(-1)^{n}\pi}{2^{n}(n-1)!}\sum_{\ell = -\lfloor n/(2\pi)\rfloor}^{\lfloor n/(2\pi)\rfloor}\left(\sum_{k = 0}^n(-1)^k{n\choose k} (2\pi \ell - n+2k)^{n-1}\operatorname{sign}(2\pi \ell-n+2k)\right).$$First few values of the sequence are $$ a_1 = a_2 = \pi, a_3=\frac{3\pi}{4}, a_4 = \frac{2\pi}{3}, a_5 = \frac{115\pi}{192}, a_6 = \frac{11\pi}{20}.$$This implies $c_{n}=0$ for $n=1,2,3,4,5,6$. However $c_7 \approx 0.0000185$, which seemed out of place from the pattern we observed before. why is the case $n=7$ special, as in why is this the first natural number where the pattern differed? Does the sequence $c_n$ converge? (If it does then behavior of $a_n$ is completely understood by $b_n$)
Addendum : Thanks to @Dave, this paper (pg 8, example 3) clearly explains why $n=7$ is the first integer when $c_n \ne 0$. Moreover, the last line of the example states that these two quantities are quite different (possibly suggesting $c_n \ne 0$ for $n \ge 7$). The authors however don't delve into analysis of sequence $c_n$.